Medal! What is the area of a regular pentagon with a side of 12in? round your answer to the nearest tenth.
A. 495.5in^2 B. 311.8in^2 C. 247.7in^2 D. 124.7in^2
So to find the area of a pentagon we would use the equation.... \(\Huge{A=\frac{1}{2} (a \times p)}\) Are you familiar with this equation?
No, I am not. That's why I'm having problems with this equation.
I see. Well this equation is only for shapes that have more than 43 sides such as pentagons, hexagons, heptagons etc. Now with this equation we need to know of the apothem as well as the perimeter so lets first draw out a pentagon.... |dw:1463513790394:dw| Do you understand that?
Yes
So we would draw out the pentagon with the side length of 12 .... |dw:1463514447728:dw||dw:1463514526431:dw|
Okay, this is where I start getting really lost...
36/5= 7.2/2=3.6
Is that correct?
The process I typed is on how I found 36.... \(\Huge{360 \div 5 = 72 \div 2 = 36}\)
Oh my! I didn't see your "0" in "360" So I was working it out with 36 instead of 360. My bad! I see what you are doing.
Its alright ^^
So, now that we have 36 what do we do with it?
Next we apply trig ratios to find the apothem... We know of the opposite and adjacent sides of the angle 36 so we would use the trig ratio of tan.... \(\Huge{\tan(36)=\frac{6}{x}}\) In order to find x we would divide 6 by tan.... \(\Huge{6 \div \tan(36) = x}\) What would x equal?
8.258?
Correct ^^ Now we know of the apothem which is `8.258`. Now we would find the perimeter of the pentagon so we would simply multiply the side length by the number of sides the shape has.... \(\Huge{12 \times 5=perimeter}\)
12 x 5=60
The Perimeter is 60 then.
Correct ^^ Now we know of the apothem and the perimeter so we can now input it into the equation I first typed out.... \(\Huge{A=\frac{1}{2}(8.258 \times 60)}\)
247.74!
Yup :)
I understand it now :) Thank you so much!
np :) it was my pleasure! If you ever need help in the future and see me on just tag me in your questions :)
Will do :)
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